Can we describe Quantum Mechanics using filters and matrices? Can mathematical filters or ultrafilters be used to predict quantum physics 'events' as accurately as using matrices like Schrodinger did? Is there a way to explain some of the predictive power of quantum mechanics with matrices on an amateur level?
 A: To quickly answer your question about the non-commutativity of matrices (and more generally operators): The uncertainty relation is entirely a consequence of non-commutativity. The uncertainty relation that you are talking about is a consequence of a much more general statement. It goes something like this...
Let $A,B$ be operators (thinking of the as matrices is fine) and let $AB - BA = C$ (operators, in general, do not commute, so this is usually nonzero). Then after a little bit of work one can find
$$ \Delta A \Delta B \geq \frac{1}{2} | \langle C \rangle | $$
where $\Delta A = \sqrt{ \langle A^2 \rangle - \langle A \rangle^2 }$. This follows from a famous inequality, Cauchy-Schwarz Inequality. 
The uncertainty relation that you are familiar with $ \Delta x \Delta p \geq \frac{ \hbar }{ 2 } $ is because $ \hat{x} \hat{p}_x - \hat{p}_x \hat{x} = i \hbar  $ (the hats distinguish operators from numbers). 
On a final note, the argument in the previous answer about calculus being a fundamental basis for exploring quantum is wildly incorrect. The best language for understanding quantum mechanics and discovering deeper truths would be Geometric Algebra (and, more generally, abstract algebra). Geometric algebra's are a extremely powerful language for doing quantum theory (and physics for that matter). I am talking about Clifford algebras, Jordan algebras, Lie algebras, and so on. There has been a massive paradigm shift in the past few decades in the setting theoretical physicists do quantum mechanics. This shift is entirely motivated by Geometric algebras. There is, arguably, new physics lying within these algebraic structures, waiting to be unpackaged by the clever physicist who realizes the correct way to apply them. This is not to say calculus is not important. It will always play a role, but it is not the proper setting for understanding quantum theory and beyond. 
I could go much further with this argument, but it would rapidly get to mathematically sophisticated. Since you are just looking for a little insight, there is no need to walk that path.
Edit: It has been a while since I wrote this and I would like to add something based on the other answers here. The fact that a quantum particle has a associated wave-function is not a fundamental tenet of quantum mechanics. The postulates of quantum mechanics are as follows


*

*States of a quantum system are described by normalized vectors in the associated Hilbert space $H$.

*For any classical observable $A$ there is a corresponding Hermitian operator  $\hat{A}$ acting in $H$. Conversely, any Hermitian operator corresponds to some observable.

*A measurement of A yields one of the eigenvalues of the corresponding operator $\hat{A}$.

*A measurement of $A$ on many identical copies of the system, all in the state $\left| \psi \rangle \right.$, produces random results. The probability to find the eigenvalue $a$ of $\hat{A}$ is $\langle \psi \left| \hat{\mathbb{I}}_a \right| \psi \rangle$, where $\hat{\mathbb{I}}_a$ is the projector onto the subspace generated by the eigenvectors associated with $a$. Equivalently, the probability density of $A$ can be written as $P(A) = \sum_a P(a) \delta (A - a) = \langle \psi \left| \delta ( A \hat{\mathbb{I}}_a - \hat{A} ) \right| \psi \rangle$.

*A measurement affects the state of the system (this is often referred to as the wave function collapse): right after the measurement the system is in the state $\left| \psi_a \rangle \right. \propto \hat{\mathbb{I}}_a \left| \psi \rangle \right.$, a normalized eigenvector of $\hat{A}$ corresponding to the eigenvalue $a$ found in the measurement.

*Between the measurements, the state of the system evolves according to the Schrodinger equation $$ i \hbar \frac{d}{dt} \left| \psi \rangle \right. = \hat{H} \left| \psi \rangle \right. $$ where $\hat{H}$ is the Hamiltonian operator of the corresponding physical system. 


That is it. All non-relativistic quantum theory stems from these axoims or, if you prefer, postulates. Further I must add that calculus is NOT what is needed to probe deeper into quantum theory. If this was true physicists would have been finished as soon as they started. The entire reason quantum mechanics is so difficult to comprehend both physically and mathematically is because the the foundation of the theory is built on mathematics that is far more sophisticated that Newton's/Leibniz's calculus. Today, most theories in modern physics do not stand on the shoulders of simple calculus, not even modern day classical mechanics! For example, the mathematical model of the Standard Model of Particle physics is group theory (extremely important, start learning it), and we know that it is even limiting us! There is an even more sophisticated foundation for the standard model called the Clifford Algebraic Standard Model, and the current hopes is that it can be the path way that help us understand the new physics scales neutrino masses are going to introduce. Now of course you we will always use calculus, but my point is that it is old news. We are far far beyond this. If you want more examples of what I am talking about look up what things like Lie theory have done for nuclear physics, what differential geometry and topology has done for General Relativity and Gravity, even look up how we formulate classical mechanics today!
A: There is always a level of mathematical understanding. The trick is to be honest with yourself on where you can understand the articulation. The most basic introductory text that describes quantum fields with mathematics( and eventually matrix formulations) that I have encountered is Griffiths - Introduction to Quantum Mechanics. It assumes some knowledge of calculus and linear algebra but that is about it.
The fundamental tenant of a quantum particle is that it behaves as a wavefunction. For this moment, just think of this wavefunction as some function on a graph like $f(x) = \cos{x}$. We can represent a quantum particle in a state $\alpha$ and a representation $q$. Usually these representations are position, momentum but they can also be more abstract concepts like energy level or spin. For example, we write $\psi_\alpha (x)$ as a wavefunction in a state $\alpha$ and at a position $x$. What we actually see are not the wavefunctions themselves but squares of the wavefunctions, namely their probability densities. We sum these wavefunctions over the particular representation in order to the find a probability of finding this particle between the a and b.
$$P_{ab} = \int\limits_a^b dx |\psi (x) |^2$$
Note that $|\psi(x)|^2 = \psi^*(x) \psi(x)$ where $\psi^*$ denotes the complex conjugate. If we want to talk about the matrix mechanics of the quantum realm we will talk about things that are now known as vectors.
That same wavefunction that I have written above? I can also write it like 
$$\psi_\alpha(x) = \langle \alpha | x \rangle $$
This is the inner product of two "vectors" $\langle \alpha |$ and $| x \rangle$. You can think of $\langle \alpha | x \rangle$ as a contraction of two vectors as a dot product into a scalar valued function.
$$\underbrace{{\boldsymbol v} \cdot {\boldsymbol u}}_{\mathrm{contracting\,\,\, two\,\,\,vectors}} = \underbrace{u_1 v_1 + u_2 v_2 + ...}_{\mathrm{scalar \,\,\,valued}}$$
This is very much the basis of the matrix formulation of quantum mechanics. Note that in reality we usually deal with infinitely dimensional vector spaces so that these arguments are just serving as an intermediate stepping stone to more abstract concepts. Most of the quantum books are not suitable for an amateur with no multivariable calculus or linear algebra background but there are some texts, like the aforementioned, that start off slow and convey important framework first. 
